What is oh in final solution prepared by mixing 20 ml of 0.05 m hcl with 30 mlof .10m barrium oxide
1 answer:
You might be interested in
Answer:
The answer is E. All of the statements describe the anomeric carbon.
Explanation:
When a sugar switches from its open form to its ring form, the carbon from the carbonyl (aldehyde if it is an aldose, or a ketone in the case of a ketose) suffers a nucleophilic addition by one of the hydroxyls in the chain, preferably one that will form a 5 or 6 membered ring after the reaction.
As such, the anomeric carbon will have two oxygens attached (The original one and the one that bonded when the ring closed).
It will be chiral, given that it has 4 different groups attached. (-OR,-OH,-H and -R, where R is the carbon chain).
The hydroxyl group can be in any position (Above of below the ring), depending on with side the addition took place. (See attachment)
It is the carbon of the carbonyl in the open-chain form of the sugar, because it is the only one that can react with the Hydroxyls.
Answer:
- Empirical:
- Molecular:
Explanation:
Hello,
In this case, based on the information regarding the combustion, the moles of carbon turn out:
Moreover, the moles of hydrogen:
Thus, the subscripts of carbon and hydrogen in the hydrocarbon turn out:
Now, looking for a suitable whole number we obtain the following empirical formula as 2.335 times 3 is 7 for hydrogen:
In such a way, that compound has a molar mass of 43 g/mol, thus, the whole compound's molar mass is 86.18 g/mol for which the molecular formula is twice the empirical one, therefore:
Which is hexane.
Best regards.